Proper Subnormal Rounding When Interpreting

During 25-bit rounding, subnormals are "normalized"
This would normally mean that the exponent needs to be able to be <-1023
Instead, you can modify at what bit you round and get the same results!
This is done by finding the highest bit and shifting right the round bit

Co-Authored-By: JosJuice <josjuice@gmail.com>

1 files changed, 27 insertions, 1 deletions

diff --git a/Source/Core/Core/PowerPC/Interpreter/Interpreter_FPUtils.h b/Source/Core/Core/PowerPC/Interpreter/Interpreter_FPUtils.h
index 2f9406ff05..67b6af7560 100644
--- a/Source/Core/Core/PowerPC/Interpreter/Interpreter_FPUtils.h
+++ b/Source/Core/Core/PowerPC/Interpreter/Interpreter_FPUtils.h
@@ -92,7 +92,33 @@ inline double Force25Bit(double d)
 {
   u64 integral = std::bit_cast<u64>(d);
 
-  integral = (integral & 0xFFFFFFFFF8000000ULL) + (integral & 0x8000000);
+  u64 exponent = integral & Common::DOUBLE_EXP;
+  u64 fraction = integral & Common::DOUBLE_FRAC;
+
+  if (exponent == 0 && fraction != 0)
+  {
+    // Subnormals get "normalized" before they're rounded
+    // In the end, this practically just means that the rounding is
+    // at a different bit
+
+    s64 keep_mask = 0xFFFFFFFFF8000000LL;
+    u64 round = 0x8000000;
+
+    // Shift the mask and rounding bit to the right until
+    // the fraction is "normal"
+    // That is to say shifting it until the MSB of the fraction
+    // would escape into the exponent
+    u32 shift = std::countl_zero(fraction) - (63 - Common::DOUBLE_FRAC_WIDTH);
+    keep_mask >>= shift;
+    round >>= shift;
+
+    // Round using these shifted values
+    integral = (integral & keep_mask) + (integral & round);
+  }
+  else
+  {
+    integral = (integral & 0xFFFFFFFFF8000000ULL) + (integral & 0x8000000);
+  }
 
   return std::bit_cast<double>(integral);
 }